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Find constants A and B such that 

 

 

\(\frac{x+7}{x^2 -x-2}=\frac{A}{x-2} + \frac{B}{x+1}\)

 

 

for all x such that \(x \ne -1\) and \(x \ne 2\). Give your answer as the ordered pair (A,B).

 

Thank you

 May 8, 2021
 #1
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If you let x go infinity, then you get 0 = A + B.

 

If you let x = 0, then you get -7/2 = A/(-2) + B.

 

The solution to this system is then A = 7/3, B = -7/3, so (A,B) = (7/3,-7/3).

 May 8, 2021
 #2
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Note that x^2-x-2   factors to   x-2   *   x+1

 

Multiply the  equation by   (x-2)(x+1)

 

x+7      =   a(x+1)   + b (x-2)

x+7   = ax + a + bx - 2b             equat the 'x' and 'non-x' components to get the following two equations

 

x = x (a+b)             and           7 = a - 2b

1 =( a+b )

a = 1-b       sub this into the red equation     7 = 1-b    - 2b       to find   b = -2    then a = 3

 May 8, 2021
 #3
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Thank you!

shananigans  May 8, 2021

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